Exam Strategy Note: Section D of the CBSE Class 12 Mathematics board exam contains long answer (LA) questions carrying 5 marks each. Mastering these standard core questions guarantees high scoring in topics including Matrices & Determinants, Calculus, Vectors & 3D Geometry, and Linear Programming.
1. Relations & Functions / Inverse Trig
Question 1 (Relations & Functions)
Let R be a relation on the set A = {1, 2, 3, 4, 5, 6, 7} given by R = {(a, b) : |a - b| \text{ is even}}. Show that R is an equivalence relation. Also, show that all elements of {1, 3, 5, 7} are related to each other and all elements of {2, 4, 6} are related to each other.
Solution:
1. Reflexivity: For any a ∈ A, |a - a| = 0, which is an even number. Thus, (a, a) ∈ R for all a ∈ A. Hence, R is reflexive.
2. Symmetry: Let (a, b) ∈ R \Rightarrow |a - b| is even. Since |a - b| = |b - a|, |b - a| is also even. Thus, (b, a) ∈ R. Hence, R is symmetric.
3. Transitivity: Let (a, b) ∈ R and (b, c) ∈ R. Then |a - b| and |b - c| are even. This implies a - b = ± 2k_1 and b - c = ± 2k_2 for integers k_1, k_2. Adding both gives a - c = ± 2(k_1 + k_2), so |a - c| is even. Thus, (a, c) ∈ R. Hence, R is transitive.
Since R is reflexive, symmetric, and transitive, R is an equivalence relation.
Part 2: Elements of {1, 3, 5, 7} are all odd, and the absolute difference between any two odd numbers is always even. Similarly, elements of {2, 4, 6} are all even, and the absolute difference between any two even numbers is always even. Thus, elements within each subset are related to each other.
Question 2 (Functions - Bijectivity)
Show that the function f : R → {x ∈ R : -1 < x < 1} defined by f(x) = \frac{x}{1 + |x|} is one-one and onto.
Solution:
1. One-One (Injective): Case (i): If x_1, x_2 ≥ 0, then f(x_1) = f(x_2) \Rightarrow \frac{x_1}{1+x_1} = \frac{x_2}{1+x_2} \Rightarrow x_1 + x_1 x_2 = x_2 + x_1 x_2 \Rightarrow x_1 = x_2.
Case (ii): If x_1, x_2 < 0, then f(x_1) = f(x_2) \Rightarrow \frac{x_1}{1-x_1} = \frac{x_2}{1-x_2} \Rightarrow x_1 - x_1 x_2 = x_2 - x_1 x_2 \Rightarrow x_1 = x_2.
Case (iii): If x_1 > 0 and x_2 < 0, f(x_1) > 0 and f(x_2) < 0, so f(x_1) ≠ f(x_2). Hence, f is one-one.
2. Onto (Surjective): Let y ∈ (-1, 1). If y ≥ 0, solve y = \frac{x}{1+x} \Rightarrow x = \frac{y}{1-y} ≥ 0. If y < 0, solve y = \frac{x}{1-x} \Rightarrow x = \frac{y}{1+y} < 0. In both cases, for every y ∈ (-1, 1), there exists x ∈ R such that f(x) = y. Hence, f is onto.
2. Matrices & Determinants
Question 3 (Matrix Inversion System)
Solve the following system of linear equations using matrix method:
2x + 3y + 3z = 5
x - 2y + z = -4
3x - y - 2z = 3
2x + 3y + 3z = 5
x - 2y + z = -4
3x - y - 2z = 3
Solution:
Write as AX = B, where:
A = [[2, 3, 3], [1, -2, 1], [3, -1, -2]], X = [[x], [y], [z]], B = [[5], [-4], [3]]
Step 1: Calculate |A| = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6) = 2(5) - 3(-5) + 3(5) = 10 + 15 + 15 = 40 ≠ 0. Thus, A^{-1} exists.
Step 2: Find Cofactors of A:
C_{11} = 5, C_{12} = 5, C_{13} = 5
C_{21} = 3, C_{22} = -13, C_{23} = 11
C_{31} = 9, C_{32} = 1, C_{33} = -7
Step 3: \text{adj}(A) = [[5, 3, 9], [5, -13, 1], [5, 11, -7]]
Step 4: X = A^{-1}B = \frac{1}{40} [[5, 3, 9], [5, -13, 1], [5, 11, -7]] [[5], [-4], [3]]
X = \frac{1}{40} [[25 - 12 + 27], [25 + 52 + 3], [25 - 44 - 21]] = \frac{1}{40} [[40], [80], [-40]] = [[1], [2], [-1]]
Answer: x = 1, y = 2, z = -1
Question 4 (Matrix Equation & Inverse)
If A = [[2, -1, 1], [-1, 2, -1], [1, -1, 2]], verify that A^3 - 6A^2 + 9A - 4I = O and hence find A^{-1}.
Solution:
Step 1: Calculate A^2 = A \cdot A = [[6, -5, 5], [-5, 6, -5], [5, -5, 6]].
Step 2: Calculate A^3 = A^2 \cdot A = [[22, -21, 21], [-21, 22, -21], [21, -21, 22]].
Step 3: Substitute into A^3 - 6A^2 + 9A - 4I:
[[22, -21, 21], [-21, 22, -21], [21, -21, 22]] - [[36, -30, 30], [-30, 36, -30], [30, -30, 36]] + [[18, -9, 9], [-9, 18, -9], [9, -9, 18]] - [[4, 0, 0], [0, 4, 0], [0, 0, 4]] = O. Verified!
Step 4: Post-multiplying by A^{-1}: A^2 - 6A + 9I - 4A^{-1} = O \Rightarrow 4A^{-1} = A^2 - 6A + 9I.
4A^{-1} = [[6, -5, 5], [-5, 6, -5], [5, -5, 6]] - [[12, -6, 6], [-6, 12, -6], [6, -6, 12]] + [[9, 0, 0], [0, 9, 0], [0, 0, 9]] = [[3, 1, -1], [1, 3, 1], [-1, 1, 3]]
A⁻¹ = 1/4 × [[3, 1, -1], [1, 3, 1], [-1, 1, 3]]
3. Continuity & Differentiability
Question 5 (Parametric Differentiation)
If x = a(\cos t + t \sin t) and y = a(\sin t - t \cos t), find \frac{d^2 y}{dx^2} at t = \frac{\pi}{4}.
Solution:
Step 1: Differentiate x with respect to t:
\frac{dx}{dt} = a(-\sin t + \sin t + t \cos t) = a t \cos t.
Step 2: Differentiate y with respect to t:
\frac{dy}{dt} = a(\cos t - (\cos t - t \sin t)) = a t \sin t.
Step 3: Find \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{at \sin t}{at \cos t} = \tan t.
Step 4: Differentiate again with respect to x:
\frac{d^2 y}{dx^2} = \frac{d}{dt}(\tan t) \cdot \frac{dt}{dx} = \sec^2 t \cdot \frac{1}{at \cos t} = \frac{\sec^3 t}{at}.
Step 5: At t = \frac{\pi}{4}: \sec(\pi/4) = \sqrt{2} \Rightarrow \frac{d^2 y}{dx^2} = \frac{(\sqrt{2})^3}{a(\pi/4)} = \frac{2\sqrt{2} \times 4}{\pi a} = \frac{8\sqrt{2}}{\pi a}.
Answer: 8√2 / (π a)
Question 6 (Logarithmic Differentiation)
Find \frac{dy}{dx} if y = (\sin x)^x + \sin^{-1}(\sqrt{x}).
Solution:
Let y = u + v, where u = (\sin x)^x and v = \sin^{-1}(\sqrt{x}). Then \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}.
Part A: For u = (\sin x)^x, take log: \ln u = x \ln(\sin x).
Differentiating w.r.t x: \frac{1}{u} \frac{du}{dx} = \ln(\sin x) + x \cdot \frac{\cos x}{\sin x} = \ln(\sin x) + x \cot x.
\frac{du}{dx} = (\sin x)^x [\ln(\sin x) + x \cot x].
Part B: For v = \sin^{-1}(\sqrt{x}), differentiating w.r.t x:
\frac{dv}{dx} = \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x(1-x)}}.
dy/dx = (sin x)ˣ [ln(sin x) + x cot x] + 1 / (2√(x - x²))
4. Applications of Derivatives
Question 7 (Maxima & Minima - Mensuration)
Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is \frac{2R}{\sqrt{3}}. Also find the maximum volume.
Solution:
Let r be the radius and h be the height of the cylinder inscribed in sphere of radius R.
From geometry: r^2 + \left(\frac{h}{2}\right)^2 = R^2 \Rightarrow r^2 = R^2 - \frac{h^2}{4}.
Volume of cylinder V = \pi r^2 h = \pi \left(R^2 - \frac{h^2}{4}\right) h = \pi \left(R^2 h - \frac{h^3}{4}\right).
Differentiating w.r.t h: \frac{dV}{dh} = \pi \left(R^2 - \frac{3h^2}{4}\right).
For maxima/minima, \frac{dV}{dh} = 0 \Rightarrow R^2 = \frac{3h^2}{4} \Rightarrow h = \frac{2R}{\sqrt{3}}.
Second Derivative: \frac{d^2 V}{dh^2} = \pi \left(-\frac{6h}{4}\right) = -\frac{3\pi h}{2} < 0 (Maximum).
Maximum Volume V_{\text{max}} = \pi \left(R^2 \cdot \frac{2R}{\sqrt{3}} - \frac{1}{4} \cdot \frac{8R^3}{3\sqrt{3}}\right) = \frac{4\pi R^3}{3\sqrt{3}}.
Height = 2R / √3, Max Volume = 4π R³ / (3√3)
Question 8 (Open Box Optimization)
An open box with a square base is to be made out of a given quantity of card board of area c^2 square units. Show that the maximum volume of the box is \frac{c^3}{6\sqrt{3}}.
Solution:
Let side of square base = x and height = y.
Surface Area S = x^2 + 4xy = c^2 \Rightarrow y = \frac{c^2 - x^2}{4x}.
Volume V = x^2 y = x^2 \left(\frac{c^2 - x^2}{4x}\right) = \frac{1}{4}(c^2 x - x^3).
Differentiating w.r.t x: \frac{dV}{dx} = \frac{1}{4}(c^2 - 3x^2). Setting \frac{dV}{dx} = 0 \Rightarrow x = \frac{c}{\sqrt{3}}.
\frac{d^2 V}{dx^2} = -\frac{6x}{4} < 0, so volume is maximum at x = \frac{c}{\sqrt{3}}.
V_{\text{max}} = \frac{1}{4} \left(c^2 \frac{c}{\sqrt{3}} - \frac{c^3}{3\sqrt{3}}\right) = \frac{1}{4} \cdot \frac{2c^3}{3\sqrt{3}} = \frac{c^3}{6\sqrt{3}}.
Proved: V_max = c³ / (6√3)
5. Integrals
Question 9 (Definite Integral Property)
Evaluate: I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx.
Solution:
Step 1: Apply property \int_0^a f(x)dx = \int_0^a f(a-x)dx:
I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} dx = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} dx.
Step 2: Add both equations:
2I = \int_{0}^{\pi} \frac{x \sin x + (\pi - x)\sin x}{1 + \cos^2 x} dx = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} dx.
Step 3: Substitute t = \cos x \Rightarrow dt = -\sin x dx. When x=0, t=1; x=\pi, t=-1.
2I = \pi \int_{1}^{-1} \frac{-dt}{1 + t^2} = \pi \int_{-1}^{1} \frac{dt}{1 + t^2} = \pi [\tan^{-1} t]_{-1}^{1} = \pi \left[\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right] = \frac{\pi^2}{2}.
I = \frac{\pi^2}{4}.
Answer: π² / 4
Question 10 (Definite Integral Log-Sine)
Evaluate: I = \int_{0}^{\pi/2} \log(\sin x) dx.
Solution:
Step 1: Using property \int_0^a f(x)dx = \int_0^a f(a-x)dx:
I = \int_{0}^{\pi/2} \log\left(\sin\left(\frac{\pi}{2} - x\right)\right) dx = \int_{0}^{\pi/2} \log(\cos x) dx.
Step 2: Adding the two integrals:
2I = \int_{0}^{\pi/2} [\log(\sin x) + \log(\cos x)] dx = \int_{0}^{\pi/2} \log(\sin x \cos x) dx = \int_{0}^{\pi/2} \log\left(\frac{\sin 2x}{2}\right) dx.
2I = \int_{0}^{\pi/2} \log(\sin 2x) dx - \int_{0}^{\pi/2} \log 2 \, dx = I_1 - \frac{\pi}{2} \log 2.
For I_1 = \int_{0}^{\pi/2} \log(\sin 2x) dx, put 2x = t \Rightarrow dx = dt/2:
I_1 = \frac{1}{2} \int_{0}^{\pi} \log(\sin t) dt = \frac{1}{2} \cdot 2 \int_{0}^{\pi/2} \log(\sin t) dt = I.
Thus, 2I = I - \frac{\pi}{2} \log 2 \Rightarrow I = -\frac{\pi}{2} \log 2.
Answer: - (π / 2) log 2
Question 11 (Partial Fractions Integration)
Evaluate: \int \frac{x^2 + 1}{(x-1)^2(x+2)} dx.
Solution:
Express in partial fractions: \frac{x^2 + 1}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}.
x^2 + 1 = A(x-1)(x+2) + B(x+2) + C(x-1)^2.
Putting x = 1 \Rightarrow 2 = 3B \Rightarrow B = 2/3.
Putting x = -2 \Rightarrow 5 = 9C \Rightarrow C = 5/9.
Comparing coefficients of x^2 \Rightarrow 1 = A + C \Rightarrow A = 1 - 5/9 = 4/9.
Integrating: \int \left( \frac{4/9}{x-1} + \frac{2/3}{(x-1)^2} + \frac{5/9}{x+2} \right) dx = \frac{4}{9}\ln|x-1| - \frac{2}{3(x-1)} + \frac{5}{9}\ln|x+2| + C.
Answer: 4/9 ln|x-1| - 2/(3(x-1)) + 5/9 ln|x+2| + C
6. Application of Integrals
Question 12 (Area between Parabola and Line)
Find the area of the region bounded by the parabola y^2 = 4ax and the line y = mx.
Solution:
Step 1: Find points of intersection: Substitute y = mx into curve: (mx)^2 = 4ax \Rightarrow m^2 x^2 - 4ax = 0 \Rightarrow x(m^2 x - 4a) = 0.
Points are x = 0 and x = \frac{4a}{m^2}.
Step 2: Set up Integral:
\text{Area} = \int_{0}^{4a/m^2} (\sqrt{4ax} - mx) dx = \left[ 2\sqrt{a} \cdot \frac{2}{3} x^{3/2} - m \frac{x^2}{2} \right]_{0}^{4a/m^2}
= \frac{4\sqrt{a}}{3} \left(\frac{4a}{m^2}\right)^{3/2} - \frac{m}{2} \left(\frac{4a}{m^2}\right)^2 = \frac{4\sqrt{a}}{3} \cdot \frac{8 a^{3/2}}{m^3} - \frac{m}{2} \cdot \frac{16 a^2}{m^4}
= \frac{32 a^2}{3 m^3} - \frac{8 a^2}{m^3} = \frac{8 a^2}{3 m^3}.
Area = 8a² / (3m³) sq. units
Question 13 (Area of Circle bounded by Lines)
Find the area of the smaller region bounded by the ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 and the line \frac{x}{a} + \frac{y}{b} = 1.
Solution:
Step 1: Equations of curves in First Quadrant:
Ellipse: y_1 = b \sqrt{1 - \frac{x^2}{a^2}} = \frac{b}{a} \sqrt{a^2 - x^2}. Line: y_2 = b \left(1 - \frac{x}{a}\right).
Limits of integration from x = 0 to x = a.
Step 2: Integration:
\text{Area} = \int_{0}^{a} (y_1 - y_2) dx = \frac{b}{a} \int_{0}^{a} \sqrt{a^2 - x^2} dx - \frac{b}{a} \int_{0}^{a} (a - x) dx
= \frac{b}{a} \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) \right]_{0}^{a} - \frac{b}{a} \left[ ax - \frac{x^2}{2} \right]_{0}^{a}
= \frac{b}{a} \left( 0 + \frac{a^2}{2} \cdot \frac{\pi}{2} \right) - \frac{b}{a} \left( a^2 - \frac{a^2}{2} \right) = \frac{\pi a b}{4} - \frac{a b}{2} = \frac{ab}{4}(\pi - 2).
Area = ab(π - 2) / 4 sq. units
7. Differential Equations
Question 14 (Linear Differential Equation)
Find the particular solution of differential equation: \frac{dy}{dx} + y \cot x = 4x \csc x, given that y = 0 when x = \frac{\pi}{2}.
Solution:
This is a linear differential equation of the form \frac{dy}{dx} + P y = Q, where P = \cot x and Q = 4x \csc x.
Integrating Factor (I.F.): I.F. = e^{\int \cot x dx} = e^{\ln(\sin x)} = \sin x.
General Solution: y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C \Rightarrow y \sin x = \int (4x \csc x) \sin x \, dx + C.
y \sin x = \int 4x dx + C \Rightarrow y \sin x = 2x^2 + C.
Apply Initial Conditions: At x = \frac{\pi}{2}, y = 0 \Rightarrow 0 \cdot \sin(\pi/2) = 2(\pi/2)^2 + C \Rightarrow C = -\frac{\pi^2}{2}.
Particular Solution: y sin x = 2x² - (π² / 2)
Question 15 (Homogeneous Differential Equation)
Solve: x \frac{dy}{dx} - y + x \sin\left(\frac{y}{x}\right) = 0.
Solution:
Rewrite as: \frac{dy}{dx} = \frac{y - x \sin(y/x)}{x} = \frac{y}{x} - \sin\left(\frac{y}{x}\right). (Homogeneous)
Substitute y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}.
v + x \frac{dv}{dx} = v - \sin v \Rightarrow x \frac{dv}{dx} = -\sin v.
Separating variables: \frac{dv}{\sin v} = -\frac{dx}{x} \Rightarrow \csc v \, dv = -\frac{dx}{x}.
Integrating: \ln|\csc v - \cot v| = -\ln|x| + \ln C = \ln\left|\frac{C}{x}\right|.
\csc(y/x) - \cot(y/x) = \frac{C}{x}.
Answer: x [csc(y/x) - cot(y/x)] = C
8. Vectors & 3D Geometry
Question 16 (Shortest Distance between Skew Lines)
Find the shortest distance between the lines \vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) and \vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k}).
Solution:
Here, \vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}, \vec{b}_1 = \hat{i} - \hat{j} + \hat{k}
\vec{a}_2 = 2\hat{i} - \hat{j} - \hat{k}, \vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k}
1. Compute \vec{a}_2 - \vec{a}_1 = \hat{i} - 3\hat{j} - 2\hat{k}.
2. Compute \vec{b}_1 \times \vec{b}_2:
\vec{b}_1 \times \vec{b}_2 = \det \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{bmatrix} = \hat{i}(-2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 + 2) = -3\hat{i} + 0\hat{j} + 3\hat{k}.
3. Magnitude: |\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + 0 + 3^2} = \sqrt{18} = 3\sqrt{2}.
4. Dot Product: (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (1)(-3) + (-3)(0) + (-2)(3) = -3 - 6 = -9.
5. Distance formula: d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \left| \frac{-9}{3\sqrt{2}} \right| = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}.
Shortest Distance = 3√2 / 2 units
Question 17 (Foot of Perpendicular & Image)
Find the foot of perpendicular and length of perpendicular from point P(1, 2, 3) to line \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}.
Solution:
Any general point Q on the line is (3\lambda + 6, 2\lambda + 7, -2\lambda + 7).
Direction ratios of PQ = (3\lambda + 5, 2\lambda + 5, -2\lambda + 4).
Since PQ is perpendicular to the given line with d.r's (3, 2, -2):
3(3\lambda + 5) + 2(2\lambda + 5) - 2(-2\lambda + 4) = 0
9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0 \Rightarrow 17\lambda + 17 = 0 \Rightarrow \lambda = -1.
Foot of perpendicular Q = (3(-1)+6, 2(-1)+7, -2(-1)+7) = (3, 5, 9).
Length PQ = \sqrt{(3-1)^2 + (5-2)^2 + (9-3)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 units.
Foot of Perpendicular = (3, 5, 9), Length = 7 units
9. Linear Programming (LPP)
Question 18 (LPP Maximization)
Solve graphically: Maximize Z = 4x + y subject to constraints: x + y \le 50, 3x + y \le 90, x \ge 0, y \ge 0.
Solution:
Step 1: Convert inequalities into equations:
Line 1: x + y = 50 passes through (50, 0) and (0, 50).
Line 2: 3x + y = 90 passes through (30, 0) and (0, 90).
Step 2: Find Corner Points: Intersection of x+y=50 and 3x+y=90 \Rightarrow 2x = 40 \Rightarrow x=20, y=30. Point B(20, 30).
Feasible Region bounded by Corner Points: O(0,0), A(30,0), B(20,30), C(0,50).
Step 3: Evaluate Z at Corner Points:
• At O(0,0): Z = 0
• At A(30,0): Z = 4(30) + 0 = 120
• At B(20,30): Z = 4(20) + 30 = 110
• At C(0,50): Z = 4(0) + 50 = 50
Maximum Z = 120 at point (30, 0)
10. Probability
Question 19 (Bayes' Theorem - Medical Test)
A doctor is to visit a patient. From past experience, probabilities that he will come by train, bus, scooter, or other means are respectively 3/10, 1/5, 1/10, and 2/5. Probabilities that he will be late are 1/4, 1/3, and 1/12 if he comes by train, bus, and scooter respectively, but if he comes by other means, he will not be late. When he arrives, he is late. What is the probability that he came by train?
Solution:
Let E_1, E_2, E_3, E_4 be events that doctor comes by train, bus, scooter, and other means.
P(E_1) = 3/10, P(E_2) = 1/5, P(E_3) = 1/10, P(E_4) = 2/5.
Let L = Event that doctor arrives late.
Given conditional probabilities: P(L|E_1) = 1/4, P(L|E_2) = 1/3, P(L|E_3) = 1/12, P(L|E_4) = 0.
By Bayes' Theorem: P(E_1|L) = \frac{P(E_1)P(L|E_1)}{\sum_{i=1}^4 P(E_i)P(L|E_i)}.
Numerator = \frac{3}{10} \times \frac{1}{4} = \frac{3}{40}.
Denominator = \left(\frac{3}{10} \times \frac{1}{4}\right) + \left(\frac{1}{5} \times \frac{1}{3}\right) + \left(\frac{1}{10} \times \frac{1}{12}\right) + 0 = \frac{3}{40} + \frac{1}{15} + \frac{1}{120} = \frac{9 + 8 + 1}{120} = \frac{18}{120} = \frac{3}{20}.
P(E_1|L) = \frac{3/40}{3/20} = \frac{3}{40} \times \frac{20}{3} = \frac{1}{2}.
Probability = 1/2
Question 20 (Bayes' Theorem - Factory Machine Output)
In a factory, machines A, B, and C produce 50%, 30%, and 20% of total items. Percentage of defective items produced are 2%, 1%, and 4% respectively. An item is chosen at random and found defective. Find probability it was produced by Machine A.
Solution:
Let E_1, E_2, E_3 be events item produced by A, B, C. P(E_1) = 0.5, P(E_2) = 0.3, P(E_3) = 0.2.
Let D be event that item is defective. P(D|E_1) = 0.02, P(D|E_2) = 0.01, P(D|E_3) = 0.04.
P(E_1|D) = \frac{P(E_1)P(D|E_1)}{P(E_1)P(D|E_1) + P(E_2)P(D|E_2) + P(E_3)P(D|E_3)}.
= \frac{0.5 \times 0.02}{(0.5 \times 0.02) + (0.3 \times 0.01) + (0.2 \times 0.04)} = \frac{0.010}{0.010 + 0.003 + 0.008} = \frac{0.010}{0.021} = \frac{10}{21}.
Probability = 10 / 21
11. Additional High-Yield Questions (21 - 30)
Question 21 (Relations - Equivalence Classes)
Show that relation R on Z \times Z defined by (a, b) R (c, d) \iff a + d = b + c is an equivalence relation.
Solution:
1. Reflexive: a + b = b + a \implies (a,b) R (a,b).
2. Symmetric: (a,b) R (c,d) \implies a+d = b+c \implies c+b = d+a \implies (c,d) R (a,b).
3. Transitive: (a,b)R(c,d) and (c,d)R(e,f) \implies a+d = b+c and c+f = d+e. Adding gives a+d+c+f = b+c+d+e \implies a+f = b+e \implies (a,b)R(e,f).
Hence, R is an Equivalence Relation.
Question 22 (Matrix Inversion Application)
If A = [[1, -1, 1], [2, 1, -3], [1, 1, 1]] and B = [[4, 2, 2], [-5, 0, 5], [1, -2, 3]], find AB and use it to solve: x - y + z = 4, 2x + y - 3z = 0, x + y + z = 2.
Solution:
AB = [[1, -1, 1], [2, 1, -3], [1, 1, 1]] [[4, 2, 2], [-5, 0, 5], [1, -2, 3]] = [[8, 0, 0], [0, 8, 0], [0, 0, 8]] = 8I.
Hence A^{-1} = \frac{1}{8} B. Given system is AX = C \implies X = A^{-1}C = \frac{1}{8} B [[4], [0], [2]].
X = \frac{1}{8} [[16 + 0 + 4], [-20 + 0 + 10], [4 + 0 + 6]] = \frac{1}{8} [[20], [-10], [10]].
Answer: x = 5/2, y = -5/4, z = 5/4
Question 23 (Second Order Derivative)
If y = ( \tan^{-1} x )^2, show that (x^2 + 1)^2 y_2 + 2x(x^2 + 1)y_1 = 2.
Solution:
Differentiating w.r.t x: y_1 = 2 \tan^{-1} x \cdot \frac{1}{1 + x^2} \implies (1 + x^2) y_1 = 2 \tan^{-1} x.
Differentiating again w.r.t x: (1 + x^2) y_2 + 2x y_1 = \frac{2}{1 + x^2}.
Multiplying both sides by (1 + x^2) gives: (x^2 + 1)^2 y_2 + 2x(x^2 + 1)y_1 = 2.
Proved!
Question 24 (Derivative of Inverse Function)
If \cos y = x \cos(a + y), with \cos a \neq \pm 1, prove that \frac{dy}{dx} = \frac{\cos^2(a + y)}{\sin a}.
Solution:
Express x = \frac{\cos y}{\cos(a + y)}. Differentiate w.r.t y:
\frac{dx}{dy} = \frac{\cos(a+y)(-\sin y) - \cos y (-\sin(a+y))}{\cos^2(a+y)} = \frac{\sin(a+y)\cos y - \cos(a+y)\sin y}{\cos^2(a+y)}.
\frac{dx}{dy} = \frac{\sin(a+y - y)}{\cos^2(a+y)} = \frac{\sin a}{\cos^2(a+y)} \implies \frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin a}.
Proved!
Question 25 (Maxima & Minima - Cone in Sphere)
Show that altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is \frac{4r}{3}.
Solution:
Let x be distance from center to base of cone. Height h = r + x, base radius R = \sqrt{r^2 - x^2}.
V = \frac{1}{3}\pi R^2 h = \frac{1}{3}\pi (r^2 - x^2)(r + x).
\frac{dV}{dx} = \frac{1}{3}\pi [(r^2 - x^2)(1) + (r + x)(-2x)] = \frac{1}{3}\pi (r + x)(r - 3x) = 0 \implies x = r/3.
Thus height h = r + r/3 = \frac{4r}{3}.
Proved: Altitude = 4r / 3
Question 26 (Integration by Parts / Special Form)
Evaluate: \int \frac{x e^x}{(x+1)^2} dx.
Solution:
Rewrite integrand: \frac{x}{(x+1)^2} e^x = \left(\frac{x+1-1}{(x+1)^2}\right) e^x = \left(\frac{1}{x+1} - \frac{1}{(x+1)^2}\right) e^x.
Matches form \int e^x [f(x) + f'(x)] dx = e^x f(x) + C, where f(x) = \frac{1}{x+1}.
Answer: eˣ / (x + 1) + C
Question 27 (Area Bounded by Circles)
Find the area of the region {(x,y) : x^2 + y^2 \le 4, x + y \ge 2}.
Solution:
Points of intersection of circle x^2 + y^2 = 4 and line x + y = 2 are (2,0) and (0,2).
\text{Area} = \int_{0}^{2} \left( \sqrt{4 - x^2} - (2 - x) \right) dx = \left[ \frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\frac{x}{2} - 2x + \frac{x^2}{2} \right]_0^2.
= \left(0 + 2\left(\frac{\pi}{2}\right) - 4 + 2\right) - 0 = \pi - 2.
Area = (π - 2) sq. units
Question 28 (Differential Equation - Integrating Factor)
Solve: (1 + x^2)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}.
Solution:
Standard form: \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{1}{(1+x^2)^2}.
I.F. = e^{\int \frac{2x}{1+x^2}dx} = e^{\ln(1+x^2)} = 1+x^2.
Solution: y(1+x^2) = \int \frac{1+x^2}{(1+x^2)^2} dx + C = \int \frac{1}{1+x^2} dx + C = \tan^{-1} x + C.
Answer: y(1 + x²) = tan⁻¹(x) + C
Question 29 (3D Geometry - Line and Plane)
Find vector equation of line passing through (1, 2, 3) and parallel to planes \vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 5 and \vec{r} \cdot (3\hat{i} + \hat{j} + \hat{k}) = 6.
Solution:
Parallel vector \vec{b} = \vec{n}_1 \times \vec{n}_2 = \det \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & 1 \end{bmatrix} = \hat{i}(-1-2) - \hat{j}(1-6) + \hat{k}(1+3) = -3\hat{i} + 5\hat{j} + 4\hat{k}.
Line passes through \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}.
Line: r̄ = (î + 2ĵ + 3k̂) + λ(-3î + 5ĵ + 4k̂)
Question 30 (LPP Minimization)
Solve graphically: Minimize Z = 200x + 500y subject to: x + 2y \ge 10, 3x + 4y \le 24, x \ge 0, y \ge 0.
Solution:
Corner Points of feasible region formed by lines x + 2y = 10 and 3x + 4y = 24 are:
A(0, 5), B(4, 3), C(0, 6) and D(8, 0).
• Z(0, 5) = 2500
• Z(4, 3) = 200(4) + 500(3) = 800 + 1500 = 2300
• Z(0, 6) = 3000
• Z(8, 0) = 1600
Minimum Z = 1600 at point (8, 0)
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